You probably squared both sides and tried to solve the resulting quadratic equation by factoring. However, the solutions to $\displaystyle{0}={\left({a}-{1}\right)}{\left({a}-{8}\right)}\usepackage{cmbright}\Huge\$ are not $\displaystyle-{1}\usepackage{cmbright}\Huge\$ and $\displaystyle-{8}\usepackage{cmbright}\Huge\$ . Remember that this equation becomes two linear equations: $\displaystyle{a}-{1}={0}\usepackage{cmbright}\Huge\$ and $\displaystyle{a}-{8}={0}\usepackage{cmbright}\Huge\$ . Solve these and then check your results back in the original equation. If a solution does not make the equation true, it is an extraneous solution. The correct answer is $\displaystyle{a}={8}\usepackage{cmbright}\Huge\$ .